For REF  and
REF , see
[1]}.
To prove REF , note that
\(\Vert  (0,(c,G)) \Vert  = \Vert  (c,G) \Vert \) 
by the very definition REF  of
\(\) , and that \(\Gamma \subseteq (0,(c,G))\) . It remains to show that
\(\Vert  (0,(c,G)) \Vert  \le \Vert  \Gamma \Vert \) . To this end, let
\(p = \alpha c + y\)  for some \(\alpha \in {-1,1}\)  and
\(y \in G {-1,1}^q\) . Then
\(p = \lambda (c+y) + ( 1 - \lambda ) (y-c)\)  for
\(\lambda = ( 1 + \alpha ) / 2 \in {0,1}\) , and so
\(\Vert  p \Vert  \le \max \lbrace  \Vert  c + y \Vert , \Vert  c - y \Vert  \rbrace \)  as \(\Vert  \cdot \Vert \)  is
convex. It follows that \(\Vert  p \Vert  \le \Vert  \Gamma \Vert \)  since
\(c+y,c-y \in \Gamma \) .
The set inclusion claim in REF  is known
[2]}; we sketch a proof: Denote
\(E = ( ( (b,F), (c,G) ) )\)  and
let \(x = b + F \lambda \)  for some \(\lambda \in {-1,1}^p\) . Then
\(x = (b+c)/2 + (b-c)/2 + (F+G) \lambda /2 + (F-G) \lambda /2 \in E\) .
This shows that \(\Omega \subseteq E\) , and similarly we obtain
\(\Gamma \subseteq E\) . The claim follows as \(E\)  is convex.
To prove the estimate, which improves Girard's result [2]},
let \(x \in E\) . Then
\(2 x=b + c+\alpha ( b - c )+( F + G ) \mu +( F - G ) \nu \) 
for some \(\alpha \in {-1,1}\)  and
some \(\mu , \nu \in {-1,1}^p\) . Define
\(y=\lambda ( c + G \mu )+ ( 1 - \lambda ) ( b + F \mu )\) 
for \(\lambda = ( 1 - \alpha ) / 2 \in {0,1}\) .
Then \(y \in ( \Omega \cup \Gamma )\)  by Lemma
REF ,
and
\(x - y=( F - G ) ( \nu - \alpha \mu ) / 2\) .
Since \(\Vert  \nu - \alpha \mu \Vert _{\infty } \le 2\) , we arrive at
\(\Vert  x - y \Vert  \le \Vert  F - G \Vert \) , which proves the bound.
